“Know how to solve every problem that has been solved.” “What I cannot create, I do not understand.” — Richard Feynman

The Particle in a Box by Hand — Degeneracy in a square box

Exercises

Introductory Quantum Mechanics › Unit 2 · Bound states in one dimension › The Particle in a Box by Hand › all problems

Practice Problem 5 of 10

Degeneracy in a square box

Problem

In a 2D square box,

E \propto n_{x}^{2} + n_{y}^{2}

. List the first three degenerate levels and their states. Then find the lowest level that is triply degenerate — where the degeneracy is NOT just the

n_{x} \leftrightarrow n_{y}

swap.

Solution

Predict before reading on. The technique is still “energies from integers,” but the counting step of the worked example becomes the whole problem. Swapping $(n_{x}, n_{y})$ always pairs states — what number-theoretic accident would produce a third state?

n_{x}^{2} + n_{y}^{2} = 5 : (1, 2), (2, 1) 10 : (1, 3), (3, 1) 13 : (2, 3), (3, 2)

n_{x}^{2} + n_{y}^{2} = 50 : (1, 7), (7, 1), (5, 5)

The pairs come from the box’s diagonal symmetry. The triple at 50 is different: $50 = 1 + 49 = 25 + 25$ is expressible as a sum of two squares in two genuinely different ways — an accidental degeneracy from number theory, not geometry. Accidental degeneracies usually flag a hidden symmetry; here the “symmetry” is arithmetic.

Result

pairs at 5, 10, 13; first triple at n_{x}^{2} + n_{y}^{2} = 50